我正在尝试使用Oracle中的LISTAGG函数。我想只获得该列的不同值。有没有一种方法可以在不创建函数或过程的情况下只获取不同的值？
col1 col2 Created_by 1 2 Smith 1 2 John 1 3 Ajay 1 4 Ram 1 5 Jack
我需要选择col1和col2的LISTAGG（不考虑第3列）。当我这样做时，我会得到LISTAGG的结果： [2,2,3,4,5]
我需要在这里删除重复的'2';我只需要col2对col1的不同值。
答案 0 :(得分：57)
19c及更高版本：
select listagg(distinct the_column, ',') within group (order by the_column)
from the_table
18c及更早版本：
select listagg(the_column, ',') within group (order by the_column)
from (
select distinct the_column
from the_table
) t
如果您需要更多列，可能会出现以下内容：
select col1, listagg(col2, ',') within group (order by col2)
from (
select col1,
col2,
row_number() over (partition by col1, col2 order by col1) as rn
from foo
order by col1,col2
)
where rn = 1
group by col1;
答案 1 :(得分：37)
以下是解决问题的方法。
select
regexp_replace(
'2,2,2.1,3,3,3,3,4,4'
,'([^,]+)(,\1)*(,|$)', '\1\3')
from dual
返回
2,2.1,3,4
答案（见下面的注释）：
select col1,
regexp_replace(
listagg(
col2 , ',') within group (order by col2) -- sorted
,'([^,]+)(,\1)*(,|$)', '\1\3') )
from tableX
where rn = 1
group by col1;
注意：以上内容适用于大多数情况 - 列表应该排序，您可能需要根据您的数据修剪所有尾随和前导空格。
如果您在组中有很多项目＆gt; 20或大字符串大小你可能遇到oracle字符串大小限制'字符串连接的结果太长'所以在每个组的成员上放一个最大数字。这只有在可以仅列出第一个成员的情况下才有效。如果你有很长的变量字符串，这可能不起作用。你将不得不进行实验。
select col1,
case
when count(col2) < 100 then
regexp_replace(
listagg(col2, ',') within group (order by col2)
,'([^,]+)(,\1)*(,|$)', '\1\3')
else
'Too many entries to list...'
end
from sometable
where rn = 1
group by col1;
希望避免oracle字符串大小限制的另一个解决方案（并非如此简单） - 字符串大小限制为4000.感谢here的user3465996帖子{/ 3}
select col1 ,
dbms_xmlgen.convert( -- HTML decode
dbms_lob.substr( -- limit size to 4000 chars
ltrim( -- remove leading commas
REGEXP_REPLACE(REPLACE(
REPLACE(
XMLAGG(
XMLELEMENT("A",col2 )
ORDER BY col2).getClobVal(),
'<A>',','),
'</A>',''),'([^,]+)(,\1)*(,|$)', '\1\3'),
','), -- remove leading XML commas ltrim
4000,1) -- limit to 4000 string size
, 1) -- HTML.decode
as col2
from sometable
where rn = 1
group by col1;
一些测试用例 - 仅供参考
regexp_replace('2,2,2.1,3,3,4,4','([^,]+)(,\1)+', '\1')
-> 2.1,3,4 Fail
regexp_replace('2 ,2 ,2.1,3 ,3 ,4 ,4 ','([^,]+)(,\1)+', '\1')
-> 2 ,2.1,3,4 Success - fixed length items
项目中包含的项目，例如。 2,21
regexp_replace('2.1,1','([^,]+)(,\1)+', '\1')
-> 2.1 Fail
regexp_replace('2 ,2 ,2.1,1 ,3 ,4 ,4 ','(^|,)(.+)(,\2)+', '\1\2')
-> 2 ,2.1,1 ,3 ,4 -- success - NEW regex
regexp_replace('a,b,b,b,b,c','(^|,)(.+)(,\2)+', '\1\2')
-> a,b,b,c fail!
v3 - 正则表达式感谢伊戈尔！适用于所有情况。
select
regexp_replace('2,2,2.1,3,3,4,4','([^,]+)(,\1)*(,|$)', '\1\3') ,
---> 2,2.1,3,4 works
regexp_replace('2.1,1','([^,]+)(,\1)*(,|$)', '\1\3'),
--> 2.1,1 works
regexp_replace('a,b,b,b,b,c','([^,]+)(,\1)*(,|$)', '\1\3')
---> a,b,c works
from dual
答案 2 :(得分：9)
您可以使用未记录的wm_concat函数。
select col1, wm_concat(distinct col2) col2_list
from tab1
group by col1;
此函数返回clob列，如果需要，可以使用dbms_lob.substr将clob转换为varchar2。
答案 3 :(得分：6)
如果想要将此转换应用于多个列，我已经扩展了a_horse_with_no_name的解决方案：
SELECT * FROM
(SELECT LISTAGG(GRADE_LEVEL, ',') within group(order by GRADE_LEVEL) "Grade Levels" FROM (select distinct GRADE_LEVEL FROM Students) t) t1,
(SELECT LISTAGG(ENROLL_STATUS, ',') within group(order by ENROLL_STATUS) "Enrollment Status" FROM (select distinct ENROLL_STATUS FROM Students) t) t2,
(SELECT LISTAGG(GENDER, ',') within group(order by GENDER) "Legal Gender Code" FROM (select distinct GENDER FROM Students) t) t3,
(SELECT LISTAGG(CITY, ',') within group(order by CITY) "City" FROM (select distinct CITY FROM Students) t) t4,
(SELECT LISTAGG(ENTRYCODE, ',') within group(order by ENTRYCODE) "Entry Code" FROM (select distinct ENTRYCODE FROM Students) t) t5,
(SELECT LISTAGG(EXITCODE, ',') within group(order by EXITCODE) "Exit Code" FROM (select distinct EXITCODE FROM Students) t) t6,
(SELECT LISTAGG(LUNCHSTATUS, ',') within group(order by LUNCHSTATUS) "Lunch Status" FROM (select distinct LUNCHSTATUS FROM Students) t) t7,
(SELECT LISTAGG(ETHNICITY, ',') within group(order by ETHNICITY) "Race Code" FROM (select distinct ETHNICITY FROM Students) t) t8,
(SELECT LISTAGG(CLASSOF, ',') within group(order by CLASSOF) "Expected Graduation Year" FROM (select distinct CLASSOF FROM Students) t) t9,
(SELECT LISTAGG(TRACK, ',') within group(order by TRACK) "Track Code" FROM (select distinct TRACK FROM Students) t) t10,
(SELECT LISTAGG(GRADREQSETID, ',') within group(order by GRADREQSETID) "Graduation ID" FROM (select distinct GRADREQSETID FROM Students) t) t11,
(SELECT LISTAGG(ENROLLMENT_SCHOOLID, ',') within group(order by ENROLLMENT_SCHOOLID) "School Key" FROM (select distinct ENROLLMENT_SCHOOLID FROM Students) t) t12,
(SELECT LISTAGG(FEDETHNICITY, ',') within group(order by FEDETHNICITY) "Federal Race Code" FROM (select distinct FEDETHNICITY FROM Students) t) t13,
(SELECT LISTAGG(SUMMERSCHOOLID, ',') within group(order by SUMMERSCHOOLID) "Summer School Key" FROM (select distinct SUMMERSCHOOLID FROM Students) t) t14,
(SELECT LISTAGG(FEDRACEDECLINE, ',') within group(order by FEDRACEDECLINE) "Student Decl to Prov Race Code" FROM (select distinct FEDRACEDECLINE FROM Students) t) t15
这是Oracle Database 11g企业版11.2.0.2.0版 - 64位生产版。
我无法使用STRAGG，因为无法进行DISTINCT和ORDER。
由于我要添加所有感兴趣的列，因此性能会线性扩展，这很好。 77K行上面花了3秒钟。只需一个汇总，.172秒。我这样做有一种方法可以在一次通过中对表中的多个列进行区分。
答案 4 :(得分：6)
我首先通过对值进行分组，然后使用listagg进行另一次聚合来克服此问题。像这样：
select a,b,listagg(c,',') within group(order by c) c, avg(d)
from (select a,b,c,avg(d)
from table
group by (a,b,c))
group by (a,b)
只有一个完整的表访问权限，相对容易扩展到更复杂的查询
答案 5 :(得分：5)
如果你想在MULTIPLE列中使用不同的值，想要控制排序顺序，不想使用可能消失的未记录的函数，并且不希望进行多次全表扫描，你会发现这个结构很有用：
with test_data as
(
select 'A' as col1, 'T_a1' as col2, '123' as col3 from dual
union select 'A', 'T_a1', '456' from dual
union select 'A', 'T_a1', '789' from dual
union select 'A', 'T_a2', '123' from dual
union select 'A', 'T_a2', '456' from dual
union select 'A', 'T_a2', '111' from dual
union select 'A', 'T_a3', '999' from dual
union select 'B', 'T_a1', '123' from dual
union select 'B', 'T_b1', '740' from dual
union select 'B', 'T_b1', '846' from dual
)
select col1
, (select listagg(column_value, ',') within group (order by column_value desc) from table(collect_col2)) as col2s
, (select listagg(column_value, ',') within group (order by column_value desc) from table(collect_col3)) as col3s
from
(
select col1
, collect(distinct col2) as collect_col2
, collect(distinct col3) as collect_col3
from test_data
group by col1
);
答案 6 :(得分：4)
要解决字符串长度问题，您可以使用类似于XMLAGG的{{1}}，但它会返回一个clob。
然后，您可以使用listagg进行解析并获取唯一值，然后使用regexp_replace将其重新转换为字符串。如果你有大量不同的值，你仍然会以这种方式耗尽空间，但在很多情况下，下面的代码应该可以工作。
您还可以更改您使用的分隔符。就我而言，我想要＆＃39; - ＆＃39;代替＆＃39;而不是＆＃39;但是你应该能够替换我的代码中的破折号，并且如果你愿意，可以使用逗号。
dbms_lob.substr()答案 7 :(得分：2)
如何创建一个能够构成“独特”部分的专用函数：
create or replace function listagg_distinct (t in str_t, sep IN VARCHAR2 DEFAULT ',')
return VARCHAR2
as
l_rc VARCHAR2(4096) := '';
begin
SELECT listagg(val, sep) WITHIN GROUP (ORDER BY 1)
INTO l_rc
FROM (SELECT DISTINCT column_value val FROM table(t));
RETURN l_rc;
end;
/
然后用它来进行聚合：
SELECT col1, listagg_distinct(cast(collect(col_2) as str_t ), ', ')
FROM your_table
GROUP BY col_1;
答案 8 :(得分：2)
即将到来的Oracle 19c将支持DISTINCT和LISTAGG。
此功能随19c一起提供：
SQL> select deptno, listagg (distinct sal,', ') within group (order by sal) 2 from scott.emp 3 group by deptno;
编辑：
LISTAGG聚合函数现在通过使用新的DISTINCT关键字来支持重复消除。 LISTAGG聚合函数根据ORDER BY表达式对查询中每个组的行进行排序，然后将这些值连接到一个字符串。使用新的DISTINCT关键字，可以在连接成单个字符串之前从指定的表达式中删除重复的值。 这消除了在使用聚合的LISTAGG函数之前创建复杂查询处理以查找不同值的需要。使用DISTINCT选项，可以直接在LISTAGG函数中完成删除重复值的处理。结果是更简单，更快，更有效的SQL。
答案 9 :(得分：1)
使用DECODE vs CASE（i saw here）进一步完善@YoYo对基于@a_horse_with_no_name的row_number（）方法的更正。我看到@Martin Vrbovsky也有此案例方法的答案。
select
col1,
listagg(col2, ',') within group (order by col2) AS col2_list,
listagg(col3, ',') within group (order by col3) AS col3_list,
SUM(col4) AS col4
from (
select
col1,
decode(row_number() over (partition by col1, col2 order by null),1,col2) as col2,
decode(row_number() over (partition by col1, col3 order by null),1,col3) as col3
from foo
)
group by col1;
答案 10 :(得分：0)
非常简单-在查询中使用具有不同选择项的子查询：
isNew()答案 11 :(得分：0)
select col1, listaggr(col2,',') within group(Order by col2) from table group by col1表示将字符串（col2）聚合到列表中，保持顺序n，然后按col1分组处理重复项，这意味着将col1重复项合并为一组。也许这看起来应该很干净简单
如果还需要col3，则只需再添加一个select col1, listaggr(col2,',') within group(Order by col2),listaggr(col3,',') within group(order by col3) from table group by col1
答案 12 :(得分：0)
如@a_horse_with_no_name所指出的那样，在调用LISTAGG之前将SELECT DISTINCT ...用作子查询的一部分可能是进行简单查询的最佳方法
但是，在更复杂的查询中，可能无法或不容易做到这一点。我是在使用带有分析功能的top-n方法的情况下提出这个问题的。
因此，我发现了COLLECT聚合函数。它被记录为具有UNIQUE或DISTINCT修饰符。仅in 10g，它静默失败（它忽略修饰符而没有错误）。但是，为了克服这个问题，我从another answer来到了这个解决方案：
SELECT
...
(
SELECT LISTAGG(v.column_value,',') WITHIN GROUP (ORDER BY v.column_value)
FROM TABLE(columns_tab) v
) AS columns,
...
FROM (
SELECT
...
SET(CAST(COLLECT(UNIQUE some_column ORDER BY some_column) AS tab_typ)) AS columns_tab,
...
)
基本上，通过使用SET，我删除了收藏集中的重复项。
您仍然需要将tab_typ定义为基本集合类型，对于VARCHAR，这将是例如：
CREATE OR REPLACE type tab_typ as table of varchar2(100)
/
作为多列情况下@a_horse_with_no_name的答案的更正，您可能希望在第三（或更多）列上进行聚合：
select
col1,
listagg(CASE rn2 WHEN 1 THEN col2 END, ',') within group (order by col2) AS col2_list,
listagg(CASE rn3 WHEN 1 THEN col3 END, ',') within group (order by col3) AS col3_list,
SUM(col4) AS col4
from (
select
col1,
col2,
row_number() over (partition by col1, col2 order by null) as rn2,
row_number() over (partition by col1, col3 order by null) as rn3
from foo
)
group by col1;
如果您将rn = 1保留为查询的条件，则将错误地汇总其他列。
答案 13 :(得分：0)
我实现了这个存储的函数：
CREATE TYPE LISTAGG_DISTINCT_PARAMS AS OBJECT (ELEMENTO VARCHAR2(2000), SEPARATORE VARCHAR2(10));
CREATE TYPE T_LISTA_ELEMENTI AS TABLE OF VARCHAR2(2000);
CREATE TYPE T_LISTAGG_DISTINCT AS OBJECT (
LISTA_ELEMENTI T_LISTA_ELEMENTI,
SEPARATORE VARCHAR2(10),
STATIC FUNCTION ODCIAGGREGATEINITIALIZE(SCTX IN OUT T_LISTAGG_DISTINCT)
RETURN NUMBER,
MEMBER FUNCTION ODCIAGGREGATEITERATE (SELF IN OUT T_LISTAGG_DISTINCT,
VALUE IN LISTAGG_DISTINCT_PARAMS )
RETURN NUMBER,
MEMBER FUNCTION ODCIAGGREGATETERMINATE (SELF IN T_LISTAGG_DISTINCT,
RETURN_VALUE OUT VARCHAR2,
FLAGS IN NUMBER )
RETURN NUMBER,
MEMBER FUNCTION ODCIAGGREGATEMERGE (SELF IN OUT T_LISTAGG_DISTINCT,
CTX2 IN T_LISTAGG_DISTINCT )
RETURN NUMBER
);
CREATE OR REPLACE TYPE BODY T_LISTAGG_DISTINCT IS
STATIC FUNCTION ODCIAGGREGATEINITIALIZE(SCTX IN OUT T_LISTAGG_DISTINCT) RETURN NUMBER IS
BEGIN
SCTX := T_LISTAGG_DISTINCT(T_LISTA_ELEMENTI() , ',');
RETURN ODCICONST.SUCCESS;
END;
MEMBER FUNCTION ODCIAGGREGATEITERATE(SELF IN OUT T_LISTAGG_DISTINCT, VALUE IN LISTAGG_DISTINCT_PARAMS) RETURN NUMBER IS
BEGIN
IF VALUE.ELEMENTO IS NOT NULL THEN
SELF.LISTA_ELEMENTI.EXTEND;
SELF.LISTA_ELEMENTI(SELF.LISTA_ELEMENTI.LAST) := TO_CHAR(VALUE.ELEMENTO);
SELF.LISTA_ELEMENTI:= SELF.LISTA_ELEMENTI MULTISET UNION DISTINCT SELF.LISTA_ELEMENTI;
SELF.SEPARATORE := VALUE.SEPARATORE;
END IF;
RETURN ODCICONST.SUCCESS;
END;
MEMBER FUNCTION ODCIAGGREGATETERMINATE(SELF IN T_LISTAGG_DISTINCT, RETURN_VALUE OUT VARCHAR2, FLAGS IN NUMBER) RETURN NUMBER IS
STRINGA_OUTPUT CLOB:='';
LISTA_OUTPUT T_LISTA_ELEMENTI;
TERMINATORE VARCHAR2(3):='...';
LUNGHEZZA_MAX NUMBER:=4000;
BEGIN
IF SELF.LISTA_ELEMENTI.EXISTS(1) THEN -- se esiste almeno un elemento nella lista
-- inizializza una nuova lista di appoggio
LISTA_OUTPUT := T_LISTA_ELEMENTI();
-- riversamento dei soli elementi in DISTINCT
LISTA_OUTPUT := SELF.LISTA_ELEMENTI MULTISET UNION DISTINCT SELF.LISTA_ELEMENTI;
-- ordinamento degli elementi
SELECT CAST(MULTISET(SELECT * FROM TABLE(LISTA_OUTPUT) ORDER BY 1 ) AS T_LISTA_ELEMENTI ) INTO LISTA_OUTPUT FROM DUAL;
-- concatenazione in una stringa
FOR I IN LISTA_OUTPUT.FIRST .. LISTA_OUTPUT.LAST - 1
LOOP
STRINGA_OUTPUT := STRINGA_OUTPUT || LISTA_OUTPUT(I) || SELF.SEPARATORE;
END LOOP;
STRINGA_OUTPUT := STRINGA_OUTPUT || LISTA_OUTPUT(LISTA_OUTPUT.LAST);
-- se la stringa supera la dimensione massima impostata, tronca e termina con un terminatore
IF LENGTH(STRINGA_OUTPUT) > LUNGHEZZA_MAX THEN
RETURN_VALUE := SUBSTR(STRINGA_OUTPUT, 0, LUNGHEZZA_MAX - LENGTH(TERMINATORE)) || TERMINATORE;
ELSE
RETURN_VALUE:=STRINGA_OUTPUT;
END IF;
ELSE -- se non esiste nessun elemento, restituisci NULL
RETURN_VALUE := NULL;
END IF;
RETURN ODCICONST.SUCCESS;
END;
MEMBER FUNCTION ODCIAGGREGATEMERGE(SELF IN OUT T_LISTAGG_DISTINCT, CTX2 IN T_LISTAGG_DISTINCT) RETURN NUMBER IS
BEGIN
RETURN ODCICONST.SUCCESS;
END;
END; -- fine corpo
CREATE
FUNCTION LISTAGG_DISTINCT (INPUT LISTAGG_DISTINCT_PARAMS) RETURN VARCHAR2
PARALLEL_ENABLE AGGREGATE USING T_LISTAGG_DISTINCT;
// Example
SELECT LISTAGG_DISTINCT(LISTAGG_DISTINCT_PARAMS(OWNER, ', ')) AS LISTA_OWNER
FROM SYS.ALL_OBJECTS;
我很抱歉，但在某些情况下（对于一个非常大的集合），Oracle可能会返回此错误：
Object or Collection value was too large. The size of the value
might have exceeded 30k in a SORT context, or the size might be
too big for available memory.
但我认为这是一个很好的开始点;）
答案 14 :(得分：0)
LISTAGG的一个令人讨厌的方面是，如果连接字符串的总长度超过4000个字符（SQL中VARCHAR2的限制），则抛出以下错误，这在Oracle版本中很难管理高达12.1
ORA-01489：字符串连接的结果太长
12cR2中添加的新功能是ON OVERFLOW的{{1}}子句。
包含此子句的查询如下所示：
LISTAGG以上将输出限制为4000个字符，但不会引发SELECT pid, LISTAGG(Desc, ' ' on overflow truncate) WITHIN GROUP (ORDER BY seq) AS desc
FROM B GROUP BY pid;
错误。
这些是ORA-01489子句的一些附加选项：
ON OVERFLOW：这会显示ON OVERFLOW TRUNCATE 'Contd..'
字符串的结尾（默认为'Contd..'）...：这将显示4000个字符
没有任何终止字符串。ON OVERFLOW TRUNCATE ''：这将显示总数
终止字符后的结尾处的字符数。
例如： - 'ON OVERFLOW TRUNCATE WITH COUNT'...(5512)：如果您希望ON OVERFLOW ERROR失败
LISTAGG错误（无论如何都是默认值）。答案 15 :(得分：0)
我找到了这个DISTINCT版本并得到了这个。
RTRIM(REGEXP_REPLACE(
(value, ', ') WITHIN GROUP( ORDER BY value)),
'([^ ]+)(, \1)+','\1'),', ')
答案 16 :(得分：0)
如果您不需要特定的连接值顺序，并且分隔符可以是逗号，则可以执行以下操作：
select col1, stragg(distinct col2)
from table
group by col1
答案 17 :(得分：0)
我写了一个函数来使用正则表达式处理它。 in参数是： 1）listagg调用本身 2）重复分隔符
create or replace function distinct_listagg
(listagg_in varchar2,
delimiter_in varchar2)
return varchar2
as
hold_result varchar2(4000);
begin
select rtrim( regexp_replace( (listagg_in)
, '([^'||delimiter_in||']*)('||
delimiter_in||'\1)+($|'||delimiter_in||')', '\1\3'), ',')
into hold_result
from dual;
return hold_result;
end;
现在您不必每次都重复正则表达式，只需说：
select distinct_listagg(
listagg(myfield,', ') within group (order by 1),
', '
)
from mytable;
答案 18 :(得分：0)
使用listagg_clob函数创建如下：
create or replace package list_const_p
is
list_sep varchar2(10) := ',';
end list_const_p;
/
sho err
create type listagg_clob_t as object(
v_liststring varchar2(32767),
v_clob clob,
v_templob number,
static function ODCIAggregateInitialize(
sctx IN OUT listagg_clob_t
) return number,
member function ODCIAggregateIterate(
self IN OUT listagg_clob_t, value IN varchar2
) return number,
member function ODCIAggregateTerminate(
self IN OUT listagg_clob_t, returnValue OUT clob, flags IN number
) return number,
member function ODCIAggregateMerge(
self IN OUT listagg_clob_t, ctx2 IN OUT listagg_clob_t
) return number
);
/
sho err
create or replace type body listagg_clob_t is
static function ODCIAggregateInitialize(sctx IN OUT listagg_clob_t)
return number is
begin
sctx := listagg_clob_t('', '', 0);
return ODCIConst.Success;
end;
member function ODCIAggregateIterate(
self IN OUT listagg_clob_t,
value IN varchar2
) return number is
begin
if nvl(lengthb(v_liststring),0) + nvl(lengthb(value),0) <= 4000 then
self.v_liststring:=self.v_liststring || value || list_const_p.list_sep;
else
if self.v_templob = 0 then
dbms_lob.createtemporary(self.v_clob, true, dbms_lob.call);
self.v_templob := 1;
end if;
dbms_lob.writeappend(self.v_clob, length(self.v_liststring), v_liststring);
self.v_liststring := value || list_const_p.list_sep;
end if;
return ODCIConst.Success;
end;
member function ODCIAggregateTerminate(
self IN OUT listagg_clob_t,
returnValue OUT clob,
flags IN number
) return number is
begin
if self.v_templob != 0 then
dbms_lob.writeappend(self.v_clob, length(self.v_liststring), self.v_liststring);
dbms_lob.trim(self.v_clob, dbms_lob.getlength(self.v_clob) - 1);
else
self.v_clob := substr(self.v_liststring, 1, length(self.v_liststring) - 1);
end if;
returnValue := self.v_clob;
return ODCIConst.Success;
end;
member function ODCIAggregateMerge(self IN OUT listagg_clob_t, ctx2 IN OUT listagg_clob_t) return number is
begin
if ctx2.v_templob != 0 then
if self.v_templob != 0 then
dbms_lob.append(self.v_clob, ctx2.v_clob);
dbms_lob.freetemporary(ctx2.v_clob);
ctx2.v_templob := 0;
else
self.v_clob := ctx2.v_clob;
self.v_templob := 1;
ctx2.v_clob := '';
ctx2.v_templob := 0;
end if;
end if;
if nvl(lengthb(self.v_liststring),0) + nvl(lengthb(ctx2.v_liststring),0) <= 4000 then
self.v_liststring := self.v_liststring || ctx2.v_liststring;
ctx2.v_liststring := '';
else
if self.v_templob = 0 then
dbms_lob.createtemporary(self.v_clob, true, dbms_lob.call);
self.v_templob := 1;
end if;
dbms_lob.writeappend(self.v_clob, length(self.v_liststring), self.v_liststring);
dbms_lob.writeappend(self.v_clob, length(ctx2.v_liststring), ctx2.v_liststring);
self.v_liststring := '';
ctx2.v_liststring := '';
end if;
return ODCIConst.Success;
end;
end;
/
sho err
CREATE or replace FUNCTION listagg_clob (input varchar2) RETURN clob
PARALLEL_ENABLE AGGREGATE USING listagg_clob_t;
/
sho err
答案 19 :(得分：0)
您可以通过RegEx替换来完成。这是一个例子：
-- Citations Per Year - Cited Publications main query. Includes list of unique associated core project numbers, ordered by core project number.
SELECT ptc.pmid AS pmid, ptc.pmc_id, ptc.pub_title AS pubtitle, ptc.author_list AS authorlist,
ptc.pub_date AS pubdate,
REGEXP_REPLACE( LISTAGG ( ppcc.admin_phs_org_code ||
TO_CHAR(ppcc.serial_num,'FM000000'), ',') WITHIN GROUP (ORDER BY ppcc.admin_phs_org_code ||
TO_CHAR(ppcc.serial_num,'FM000000')),
'(^|,)(.+)(,\2)+', '\1\2')
AS projectNum
FROM publication_total_citations ptc
JOIN proj_paper_citation_counts ppcc
ON ptc.pmid = ppcc.pmid
AND ppcc.citation_year = 2013
JOIN user_appls ua
ON ppcc.admin_phs_org_code = ua.admin_phs_org_code
AND ppcc.serial_num = ua.serial_num
AND ua.login_id = 'EVANSF'
GROUP BY ptc.pmid, ptc.pmc_id, ptc.pub_title, ptc.author_list, ptc.pub_date
ORDER BY pmid;
答案 20 :(得分：0)
listagg（）忽略NULL值，因此在第一步中你可以使用lag（）函数来分析前一条记录是否具有相同的值，如果是，则为NULL，否则为“new value＆＃39;”。
WITH tab AS
(
SELECT 1 as col1, 2 as col2, 'Smith' as created_by FROM dual
UNION ALL SELECT 1 as col1, 2 as col2, 'John' as created_by FROM dual
UNION ALL SELECT 1 as col1, 3 as col2, 'Ajay' as created_by FROM dual
UNION ALL SELECT 1 as col1, 4 as col2, 'Ram' as created_by FROM dual
UNION ALL SELECT 1 as col1, 5 as col2, 'Jack' as created_by FROM dual
)
SELECT col1
, CASE
WHEN lag(col2) OVER (ORDER BY col2) = col2 THEN
NULL
ELSE
col2
END as col2_with_nulls
, created_by
FROM tab;
结果
COL1 COL2_WITH_NULLS CREAT
---------- --------------- -----
1 2 Smith
1 John
1 3 Ajay
1 4 Ram
1 5 Jack
请注意，第二个2将替换为NULL。现在你可以用它周围的listagg（）包装一个SELECT。
WITH tab AS
(
SELECT 1 as col1, 2 as col2, 'Smith' as created_by FROM dual
UNION ALL SELECT 1 as col1, 2 as col2, 'John' as created_by FROM dual
UNION ALL SELECT 1 as col1, 3 as col2, 'Ajay' as created_by FROM dual
UNION ALL SELECT 1 as col1, 4 as col2, 'Ram' as created_by FROM dual
UNION ALL SELECT 1 as col1, 5 as col2, 'Jack' as created_by FROM dual
)
SELECT listagg(col2_with_nulls, ',') WITHIN GROUP (ORDER BY col2_with_nulls) col2_list
FROM ( SELECT col1
, CASE WHEN lag(col2) OVER (ORDER BY col2) = col2 THEN NULL ELSE col2 END as col2_with_nulls
, created_by
FROM tab );
结果
COL2_LIST
---------
2,3,4,5
您也可以在多个列上执行此操作。
WITH tab AS
(
SELECT 1 as col1, 2 as col2, 'Smith' as created_by FROM dual
UNION ALL SELECT 1 as col1, 2 as col2, 'John' as created_by FROM dual
UNION ALL SELECT 1 as col1, 3 as col2, 'Ajay' as created_by FROM dual
UNION ALL SELECT 1 as col1, 4 as col2, 'Ram' as created_by FROM dual
UNION ALL SELECT 1 as col1, 5 as col2, 'Jack' as created_by FROM dual
)
SELECT listagg(col1_with_nulls, ',') WITHIN GROUP (ORDER BY col1_with_nulls) col1_list
, listagg(col2_with_nulls, ',') WITHIN GROUP (ORDER BY col2_with_nulls) col2_list
, listagg(created_by, ',') WITHIN GROUP (ORDER BY created_by) created_by_list
FROM ( SELECT CASE WHEN lag(col1) OVER (ORDER BY col1) = col1 THEN NULL ELSE col1 END as col1_with_nulls
, CASE WHEN lag(col2) OVER (ORDER BY col2) = col2 THEN NULL ELSE col2 END as col2_with_nulls
, created_by
FROM tab );
结果
COL1_LIST COL2_LIST CREATED_BY_LIST
--------- --------- -------------------------
1 2,3,4,5 Ajay,Jack,John,Ram,Smith
答案 21 :(得分：0)
我认为这可能会有所帮助 - 如果列重复，则将列值设为NULL - 然后它不会附加到LISTAGG字符串：
with test_data as
(
select 1 as col1, 2 as col2, 'Smith' as created_by from dual
union select 1, 2, 'John' from dual
union select 1, 3, 'Ajay' from dual
union select 1, 4, 'Ram' from dual
union select 1, 5, 'Jack' from dual
union select 2, 5, 'Smith' from dual
union select 2, 6, 'John' from dual
union select 2, 6, 'Ajay' from dual
union select 2, 6, 'Ram' from dual
union select 2, 7, 'Jack' from dual
)
SELECT col1 ,
listagg(col2 , ',') within group (order by col2 ASC) AS orig_value,
listagg(CASE WHEN rwn=1 THEN col2 END , ',') within group (order by col2 ASC) AS distinct_value
from
(
select row_number() over (partition by col1,col2 order by 1) as rwn,
a.*
from test_data a
) a
GROUP BY col1
结果：
COL1 ORIG DISTINCT
1 2,2,3,4,5 2,3,4,5
2 5,6,6,6,7 5,6,7
答案 22 :(得分：0)
有没有人想过使用PARTITION BY子句？在此查询中，它对我有用，可以获取应用程序服务和访问的列表。
SELECT DISTINCT T.APP_SVC_ID,
LISTAGG(RTRIM(T.ACCESS_MODE), ',') WITHIN GROUP(ORDER BY T.ACCESS_MODE) OVER(PARTITION BY T.APP_SVC_ID) AS ACCESS_MODE
FROM APP_SVC_ACCESS_CNTL T
GROUP BY T.ACCESS_MODE, T.APP_SVC_ID
我不得不删除NDA的where子句，但是你明白了。
答案 23 :(得分：-1)
处理多个listagg的最简单方法是每列使用1个WITH（子查询因子），其中包含select列中该列的listagg：
SelectedClient给出了：
WITH tab AS
(
SELECT 1 as col1, 2 as col2, 3 as col3, 'Smith' as created_by FROM dual
UNION ALL SELECT 1 as col1, 2 as col2, 3 as col3,'John' as created_by FROM dual
UNION ALL SELECT 1 as col1, 3 as col2, 4 as col3,'Ajay' as created_by FROM dual
UNION ALL SELECT 1 as col1, 4 as col2, 4 as col3,'Ram' as created_by FROM dual
UNION ALL SELECT 1 as col1, 5 as col2, 6 as col3,'Jack' as created_by FROM dual
)
, getCol2 AS
(
SELECT DISTINCT col1, listagg(col2,',') within group (order by col2) over (partition by col1) AS col2List
FROM ( SELECT DISTINCT col1,col2 FROM tab)
)
, getCol3 AS
(
SELECT DISTINCT col1, listagg(col3,',') within group (order by col3) over (partition by col1) AS col3List
FROM ( SELECT DISTINCT col1,col3 FROM tab)
)
select col1,col2List,col3List
FROM getCol2
JOIN getCol3
using (col1)